CodeIgniter: Populating a select input/ form dropd

问题:

I'm trying to populate a select input/ form dropdown with the data from the table.

HTML

<select id="room_type" name="inputInfo" >
    <option value="<?php echo set_value('room_type', '$data_room_type'); ?>"></option>
</select>

Model:

function get_room_details($data_room_type) { // line 19
    $data_room_type = array();
    $this->db->select('room_type', 'default_price');
    $this->db->from('room_type');
    $query = $this->db->get();

    if ($query->num_rows() > 0) {
        foreach ($query->result_array() as $row) {
            $data_room_type[] = $row;
        }
    }
    return $data_room_type;
}

Controller:

function index() {
    $this->load->view('/main/new_reservation');
    $this->load->model('reservations_model');
    $this->reservations_model->get_room_details($data_room_type);
}

But I received this error:

A PHP Error was encountered

Severity: Notice

Message: Undefined variable: data_room_type

Filename: controllers/reservations.php

Line Number: 8

Line number 8 is::

$this->reservations_model->get_room_details($data_room_type);

回答1:

return $query->result_array(); line will populate more than one row for you.(if you have more than one record in room_type table)

so you need to run a foreach loop in view.for normal html

<select name="">
<?php
foreach($room_type as $each)
{
    ?>
    <option value="<?=$each['rt_name']?>"><?=$each['rt_name']?></option>
    <?php
}
?>
</select>

use normal html otherwise you need to change the return data what your model returning.

May be it will help you.Please let me know if you face any problem.

回答2:

You have the syntax error means you have make something wrong. and as I can see that you have write the spelling of function wrong in model.

It is fucntion you have write and it should be function so change there and enjoy.

回答3:

fucntion get_room_details($data_room_type) {

typo function it should be

function get_room_details($data_room_type) {

$data_room_type[] = $row. Missing semicolon

It should be $data_room_type[] = $row;

You can do like this In Model : function get_room_details($data_room_type = array()) { In Controller : $this->reservations_model->get_room_details();

回答4:

This almost worked:

Model:

function pop_room_type() {
    $this->db->select('rt_name')->from('room_type');
    $query=$this->db->get();
    return $query->result_array();
}

Controller:

function index() {
    $this->load->model('reservations_model');
    $data['room_type'] = $this->reservations_model->pop_room_type();
    //echo "<pre>";print_r($data['room_type']);echo"</pre>";
    $this->load->view('/main/new_reservation', $data);
}

View:

<?php 
//$js = 'name='$room_type'';
echo form_dropdown('room_type', $room_type);
?>   

However, there's a one BIG issue whereas the "name" of all select options are the same (rt_name: as in the table's column).